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-{
- "metadata": {
- "name": "",
- "signature": "sha256:7315f4cd0dbe1ea8c121564011c9ce9945618f308024d617157c5ed6076bc7d9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter15:Composite Materials and Ceramics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex15.1:pg-299"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 15.1 : colume ratio of aluminium and boron\n",
- " \n",
- "import numpy.linalg as lin\n",
- "yal=715;# in GN/,**2\n",
- "yfe=210;# in GN/,**2\n",
- "yb=440;# in GN/,**2\n",
- "A=numpy.array([[71, 71],[71, 440]]);#\n",
- "B=numpy.array([71,210]);#\n",
- "X=lin.solve(A,B)\n",
- "print round(X[0],3),\"is volume ratio of aluminium \"\n",
- "print round(X[1],3),\"is volume ratio of boron \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.623 is volume ratio of aluminium \n",
- "0.377 is volume ratio of boron \n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file